Justifications for the index upper lower labels in tensor component transformations

Question

A (1,0)-type tensor may be written as $$ V = V^{\mu} e_{\mu} $$ The component transforms as $$ V^{\nu} = A^{\nu}{}_{\mu^\prime} V^{\mu^\prime} $$ (the basis can transform similarly) My question is, what is the justification that the label of $A$ is one upper script and one lower script? In Wald's General relativity p26, the author showed

$$ V^{\mu^\prime} = \sum_{\mu=1}^{n} V^{\mu} \frac{ \partial x^{\prime\mu^\prime} }{\partial x^{\mu}} \qquad\qquad(2.3.6) $$

The term $\frac{ \partial x^{\prime\mu^\prime} }{\partial x^{\mu}} $ can justify the label should be one upper and one lower. But, the approach in Wald's book is based on the geometry aspect of tensor.

A tensor may be defined in more general content, just an element in vector space products. Vector space is something that satisfies a couple of axioms. By this means, how to justify the term in basis transformation has the structure of upper/lower indices? Should I regard basis transformation as a tensor contraction between (1,1) and (1,0) to (1,0) something like that? Or the index positions of transormation matrix are not always strict?

Answer 1

Where you place the indices of a transformation matrix depends on what kind of object it transforms into what other kind of object. For example, let $g^{\mu\nu} = g_{\mu\nu}{}^1$ be the components of the metric tensor $$ g=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}~, \tag{1} $$ which has the property that it maps the covariant position vector $x_\mu = (ct,-\vec x)_\mu$ onto the contravariant one, $$ x^\mu = g^{\mu\nu} x_\nu =(ct,\vec x)^\mu~, $$ and thus can be used to "pull indices up or down". Multiplying the above equation with $g_{\rho\mu}$ yields $$ x_\rho=g_{\rho\mu} x^\mu = g_{\rho \mu}g^{\mu\nu} x_\nu~, $$ so $g_{\rho}^{~~~\mu} = g_{\rho\mu} g^{\mu\nu}$ is the $\rho$-$\nu$-component of the unit matrix $g^2$.

Likewise, your example $V'^\mu = A^\mu_{~~~\nu} V^\nu$ can be multiplied with $g_{\rho \mu}$ to obtain $$ V'_{\rho} = g_{\rho\mu} V'^\mu = g_{\rho\mu} A^\mu_{~~~\nu} V^\nu = A_{\rho\nu} V^\nu~. $$ This means, $A^\mu_{~~~\nu}$ maps a contravariant vector to another contravariant vector, while $A_{\mu\nu}$ maps a contravariant vector to a covariant vector. Which one you use depends on what vectors you have or need and you can easily convert between the variants by multipliying equations with $g^{\mu\nu}$ or $g_{\mu\nu}$.

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${}^1$ Strictly speaking, this is wrong, because free indices on the left- and righthand side of an equation have to match. What I wanted to express is simply, that the co- and contravariant forms of the metric tensor are identical.

Answer 2

As you say, the components of the vector $V$ transform as $$\tag{1} V^\mu={{A^\mu}_{\mu'}}V^{\mu'}. $$ Likewise, the basis vectors transform as $$\tag{2} \boldsymbol{e}_\mu={{A_\mu}^{\mu'}}\boldsymbol{e}_{\mu'}\,. $$ where ${A_\mu}^{\mu'}$ is the inverse of ${{A^\mu}_{\mu'}}\,.$ That's the reason why it is common to call the transformaton (1) contravariant and (2) covariant. The position of the index (upper or lower) is a convenient way of underpinning this in the notation.

There is nothing that forbids a convention by which all indices are either upper or lower as long as one knows what a component of a vector is and what a basis vector is. It is also not forbidden to reverse the common convention, or to call (1) covariant and (2) contravariant. But let's forget about this it makes everyones life just harder.